A train is travelling at $180 \text { km/h }$, $500\text { m }$ away from a train station, what is the constant deceleration needed to get to a complete stop at the station.

A continued question regarding the use of series and sequences to work this out instead of physics. This was in my exam and my teacher said that "Physics was unnecessary for the question." How would you do this question using simple maths?


Speed 50 m/s.

According to galileo's law of odd numbers, in an de-accelerated motion distance covered in every following second is proportional to the decreasing ratio of odd numbers.

Let it reaches the station in $t$ seconds.

So, distance covered in first second is 50 metre. using the law the distance covered in the $r^{th}$ second will be, $\frac{2(t-r)+1}{2t+1}50$.

Total distance covered will be, $500= \sum_{r=1}^t(\frac{2(t-r)+1}{2t+1})50$

which gives $t=10+\sqrt{110} \approx 20$,

$a=\frac{\Delta v}{t}=\frac{-50}{20}=\frac{-5}{2}$, so the acceleration is $-2.5\frac{m}{s^2}$.

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    $\begingroup$ The distance covered in the first second is not $50$ meters because the train is decelerating. The deceleration time is exactly $20$ seconds, not approximately. $\endgroup$ – Ross Millikan Dec 16 '14 at 15:50
  • $\begingroup$ speed is defined as distance per second. So the distance is defined as speed times a second. or $v=\frac{d}{t}$ and $d=vt$ $\endgroup$ – Dheeraj Kumar Dec 16 '14 at 15:52
  • $\begingroup$ But if the speed is changing during the second, which it is, the distance covered is not the starting speed times one second. The correct distance is $s(t)=s(0)+v(0)t+\frac 12at^2$ $\endgroup$ – Ross Millikan Dec 16 '14 at 15:54
  • $\begingroup$ well.. the actual distance covered in the first second is 48.75 but that approximation can be taken because we can again approximate the answer in the end. $\endgroup$ – Dheeraj Kumar Dec 16 '14 at 16:02
  • $\begingroup$ Yes, but you can do it without approximation at all. If the deceleration were even faster, the error would be larger. $\endgroup$ – Ross Millikan Dec 16 '14 at 16:04

If you know that the velocity is zero at the end, you know what the average velocity is. Using that and the distance, you can compute the total time. Then with the time and change in velocity, you can compute the acceleration(negative).

Not sure if this is what the teacher had in mind. Probably not if it involved series and sequences.

ETA: Limit method wasn't as simple as I thought it would be. Hopefully somebody smarter than me will take it up, but here's what I had. Also it involves Distance=Velocity$\times$Time. Not sure if that is "physics"

Imagine breaking up the trip into $x$ segments and each segment, we reduce the velocity by $\frac{180}{x}$. We can then setup an equation like $$0.5=\lim_{x\to\infty}\sum_{i=0}^x(180-\frac{i}{x}180)(\frac{t}{x})\\0.5=\lim_{x\to\infty}\sum_{i=0}^x\frac{180t(x-i)}{x^2}\\0.5=90t\\t=\frac{1}{180}hrs$$

But given the time and change in velocity, I'm still not sure how you would get the deceleration without using a physics formula.

  • $\begingroup$ Physics was unnecessary for the question. How the examiner wished to express the question, probably due to the possibility that they wanted you go in a particular direction. ~What my teacher said to me $\endgroup$ – user145591 Dec 16 '14 at 3:21
  • $\begingroup$ Are you using limits in the class? Could set up a method that way too. $\endgroup$ – turkeyhundt Dec 16 '14 at 3:24
  • $\begingroup$ Yes, can you please show me $\endgroup$ – user145591 Dec 16 '14 at 3:29
  • $\begingroup$ We have only done preliminary work on limits, can you explain how the limit changed to 90t (line 2 to line 3?) $\endgroup$ – user145591 Dec 16 '14 at 4:09
  • $\begingroup$ I can't. I used a calculator. Hopefully somebody else will be able to better answer your question. $\endgroup$ – turkeyhundt Dec 16 '14 at 4:12

Let $s(t)$ be the position in km, $v(t)$ the speed in km/hr, $a$ the acceleration in km/hr^2. We are given $s(0)=0, v(0)=180, s(t)=\frac 12, v(t)=0$ Then $$s(t)=\frac 12at^2+v(0)t+s(0)\\ v=v(0)+at\\a=-\frac {180}t\\\frac 12=-\frac 12\cdot 180t+180t=90t\\ t=\frac 1{180}\\a=-32400$$

  • 1
    $\begingroup$ If you want to use kinematics (which you are not supposed to do) you could have used this, $v^2=u^2-2as$ where v=0 because train stops, which directly gives $-a=-\frac{50\cdot50}{2\cdot500}$=-2.5, answer in one step!!! $\endgroup$ – Dheeraj Kumar Dec 16 '14 at 16:18

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